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B
secA+cosecA
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C
sinA
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D
cosA
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Solution
The correct option is BsecA+cosecA The value of (sinA+cosA)(tanA+cotA) is =(sinA+cosA)(sinAcosA+cosAsinA) =(sinA+cosA)(sin2A+cos2AcosAsinA) =sinA+cosAcosAsinA =sinAcosAsinA+cosAcosAsinA =1cosA+1sinA